# Closed formula for a certain infinite series

I came across this problem while doing some simplifications.

QUESTION. Is there a closed formula for the evaluation of this series? $$\sum_{(a,b)=1}\frac{\cos\left(\frac{a}b\right)}{a^2b^2}$$ where the sum runs over all pairs of positive integers that are relatively prime.

• @user64494 why? "infinite sum" is a correct math term, and refers to the sum of a family of numbers, which is the present case. "Sum of a series" is a correct term, yet not applicable here, since the OP's one is not a series. Aug 8 at 17:53
• @user64494 : As Pietro Majer said, "infinite sum" is a correct term, and $\sum_{x\in X}f(x)$ can be defined for any (say) real-valued function $f$ on any set $X$. One of a number of mutually equivalent ways to do that is as follows: $\sum_{x\in X}f(x):=\int_X f\,d\nu$, where $\nu$ is the counting measure on (the $\sigma$-algebra of all subsets of) $X$, if the integral exists. Aug 8 at 19:07
• @T. Amdeberhan never follow suggestions from anonymous sources :) Aug 8 at 23:20
• This question on MathStackExchange asked a similar question: math.stackexchange.com/questions/452028/… The method of solution given there in the accepted answer can be easily applied to the OPs case, similar approach using Mobius transformation. Aug 9 at 7:25
• @user64494 On the other hand, I think we all agree that an "infinitite sum" is an incorrect math term :-) Aug 9 at 8:50

Apply Möbius summation, the formula for $$\sum_{n>=1}\cos(2\pi n x)/n^2$$ to obtain:

$$11/4-45\zeta(3)/\pi^3=1.00543...\;$$

• GH from MO has given much detail in the answer below. Let me just expand on what Henri Cohen referred to as Möbius summation: $\sum_{(a,b)=1}\frac{\cos(a/b)}{a^2b^2}=\sum_{m,n}\sum_{d\vert(m,n)}\frac{\mu(d)}{m^2n^2}=\sum_k\frac{\mu(k)}{k^4}\sum_{x,y}\frac1{x^2y^2}=\frac1{\zeta(4)}\sum_{x,y}\frac1{x^2y^2}$. Aug 9 at 13:22
• Thank you for a cute answer, Henri Cohen. Aug 9 at 13:24

Let me add the details for Henri Cohen's nice answer, without claiming any originality. We have $$\sum_{m,n\geq 1}\frac{\cos\left(\frac{m}n\right)}{m^2n^2}=\zeta(4)\sum_{(a,b)=1}\frac{\cos\left(\frac{a}b\right)}{a^2b^2}.$$ On the other hand, by the identity (cf. Ron Gordon's answer here) $$\sum_{m\geq 1}\frac{\cos(mx)}{m^2}=\frac{x^2}{4}-\frac{\pi x}{2}+\zeta(2),\qquad 0\leq x\leq 2\pi,$$ we also have $$\sum_{m,n\geq 1}\frac{\cos\left(\frac{m}n\right)}{m^2n^2}=\sum_{n\geq 1}\frac{1}{n^2}\left(\frac{n^{-2}}{4}-\frac{\pi n^{-1}}{2}+\zeta(2)\right)=\frac{\zeta(4)}{4}-\frac{\pi\zeta(3)}{2}+\zeta(2)^2.$$ Comparing the right hand sides of the first and third equation, we conclude that $$\sum_{(a,b)=1}\frac{\cos\left(\frac{a}b\right)}{a^2b^2}=\frac{1}{4}-\frac{\pi\zeta(3)}{2\zeta(4)}+\frac{\zeta(2)^2}{\zeta(4)}.$$ Here we have $$\zeta(2)=\pi^2/6$$ and $$\zeta(4)=\pi^4/90$$, therefore $$\sum_{(a,b)=1}\frac{\cos\left(\frac{a}b\right)}{a^2b^2}=\frac{11}{4}-\frac{45\zeta(3)}{\pi^3}.$$

• Could you give a reference (or derivation) for the identity in the second display? Aug 8 at 21:44
• And a reference for the first identity, please. Aug 8 at 21:47
• @JoshuaStucky: Reference added. Aug 8 at 21:59
• @IosifPinelis: The first identity follows readily by grouping the pairs $(m,n)$ according to their greatest common divisor. That is, we write $m=ad$, $n=bd$ with $d=\gcd(m,n)$ and $\gcd(a,b)=1$. Aug 8 at 22:01
• @GHfromMO: thank you for furnishing the details. Aug 9 at 13:23